Answer:
[tex]v=\frac{0,25}{\pi } ft/min[/tex]
Step-by-step explanation:
There is a relation with the initial and final dimensions of the hopper according to the sand level when its high is 6 feet as follows:
[tex]\frac{h_{1} }{h_{2} }=\frac{r_{1} }{r_2}\\{r_2}=\frac{12ft*6ft}{18ft}\\ r_2=4ft[/tex]
Where the calculated [tex]r_{2}[/tex] is given with the 6 feet hgh.
Then we have the sand flow formula which is:
[tex]Q=A*v[/tex]
Where A represents the area of the transversal section and v the velocity that we need to know, the area is:
[tex]A=\pi r^{2}\\ A=\pi *4^{2}\\ A=16\pi ft^{2}[/tex]
And finally the sand is dropping when the level is 6 feet high with the velocity (v) :
[tex]v=\frac{Q}{A} \\v=\frac{4}{16\pi }[/tex]
[tex]v=\frac{0.25}{\pi } ft/min[/tex]
